Question

Ajutati-ma va rog!
Utilizand definitia derivatei sa se scrie ecuatia tangentei la graficul functiei f:R->R, f(x)= x^2 +x, in x0=1.

Answer 1

tg: y-y0=m(x-x0)
y0=f(x0)=f(1)=1^2+1=1+1=2
m=f'(x0)=f'(1)
f'(x)=(x^2+x)'=2x+1
m=f'(1)=2*1+1=3
tg: y-2=3(x-1)
tg: y-2=3x-3
tg: y-2-3x+3=0
tg: y-3x+1=0
tg:3x-y-1=0

Answer 2

$f:\mathbb{R} \rightarrow \mathbb{R},\quad f(x) = x^2+x\\ \\ x_0 = 1 \\ \\ f'(x) = 2x+1 \\ \\ \text{Ecuatia tangentei in punctul } x_0 = 1 \text{ este: }\\y - f(x_0) = f'(x_0)(x-x_0) \\ \\ y - (1^2+1) = (2\cdot 1+1)(x-1) \\ \\y - 2 = 3(x-1) \\ \\ y-2 = 3x-3 \\ \\ \Rightarrow 3x-y -1 = 0$